Answer to Question #229223 in Chemical Engineering for Lokika

Question

Solve dy/dx= 4x² (y-x)² + if y /x is a particular solution.

Answer 1

Given DE:

$dy/dx= 4x² (y-x)² +y/x$

Let

$u(x)=y/x$

Then

$y=u\cdot x,\quad y'=u'x+u$

We get

$u'x+u=4x^2(ux-x)^2+u$ $u'x=4x^4(u-1)^2$

$\frac{du}{(u-1)^2}=4x^3dx$

$-\frac{1}{u-1}=x^4+C$

$u-1=-\frac{1}{x^4+C}$

$u=1-\frac{1}{x^4+C}$

Finally

$y(x)=x\left(1-\frac{1}{x^4+C}\right)$